Simplify the following expression: $y = \dfrac{-7x^2- 23x+20}{-7x + 5}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-7)}{(20)} &=& -140 \\ {a} + {b} &=& &=& {-23} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-140$ and add them together. Remember, since $-140$ is negative, one of the factors must be negative. The factors that add up to ${-23}$ will be your ${a}$ and ${b}$ When ${a}$ is ${5}$ and ${b}$ is ${-28}$ $ \begin{eqnarray} {ab} &=& ({5})({-28}) &=& -140 \\ {a} + {b} &=& {5} + {-28} &=& -23 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-7}x^2 +{5}x) + ({-28}x +{20}) $ Factor out the common factors: $ x(-7x + 5) + 4(-7x + 5)$ Now factor out $(-7x + 5)$ $ (-7x + 5)(x + 4)$ The original expression can therefore be written: $ \dfrac{(-7x + 5)(x + 4)}{-7x + 5}$ We are dividing by $-7x + 5$ , so $-7x + 5 \neq 0$ Therefore, $x \neq \frac{5}{7}$ This leaves us with $x + 4; x \neq \frac{5}{7}$.